A) \[-\frac{\pi }{2},\,\frac{\pi }{2}\]
B) \[-\frac{{{\pi }^{3}}}{8},\,\frac{{{\pi }^{3}}}{8}\]
C) \[\frac{7{{\pi }^{3}}}{8},\,\,\frac{{{\pi }^{3}}}{32}\]
D) None of these
Correct Answer: C
Solution :
We have \[{{({{\sin }^{-1}}x)}^{3}}+{{({{\cos }^{-1}}x)}^{3}}\] \[={{({{\sin }^{-1}}x+{{\cos }^{-1}}x)}^{3}}\]\[-3{{\sin }^{-1}}x{{\cos }^{-1}}x({{\sin }^{-1}}x+{{\cos }^{-1}}x)\] =\[\frac{{{\pi }^{3}}}{8}-3({{\sin }^{-1}}x{{\cos }^{-1}}x)\frac{\pi }{2}\] =\[\frac{{{\pi }^{3}}}{8}-\frac{3\pi }{2}{{\sin }^{-1}}x\left( \frac{\pi }{2}-{{\sin }^{-1}}x \right)\] = \[\frac{{{\pi }^{3}}}{8}-\frac{3{{\pi }^{2}}}{4}{{\sin }^{-1}}x+\frac{3\pi }{2}{{({{\sin }^{-1}}x)}^{2}}\] =\[\frac{{{\pi }^{3}}}{8}+\frac{3\pi }{2}\left[ {{({{\sin }^{-1}}x)}^{2}}-\frac{\pi }{2}{{\sin }^{-1}}x \right]\] \[=\frac{{{\pi }^{3}}}{8}+\frac{3\pi }{2}\left[ {{\left( {{\sin }^{-1}}x-\frac{\pi }{4} \right)}^{2}} \right]-\frac{3{{\pi }^{3}}}{32}\] \[=\frac{{{\pi }^{3}}}{32}+\frac{3\pi }{2}{{\left( {{\sin }^{-1}}x-\frac{\pi }{4} \right)}^{2}}\] \ The least value is \[\frac{{{\pi }^{3}}}{32}\] and since \[{{\left( {{\sin }^{-1}}x-\frac{\pi }{4} \right)}^{2}}\le {{\left( \frac{3\pi }{4} \right)}^{2}}\] \ The greatest value is \[\frac{{{\pi }^{3}}}{32}+\frac{9{{\pi }^{2}}}{16}\times \frac{3\pi }{2}=\frac{7{{\pi }^{3}}}{8}\].
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